AB is a segment, the point P is on the perpendicular bisector on segment AB such that the length of AP exceeds length of AB by 7cm. if the perimeter of ABP is 38cm. find the sides of triangle ABP
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Given:-
- AB is a segment, the point P is the perpendicular bisector of segment AB.
- The length of the AP exceeds the length of AB by 7cm.
- The perimeter of ∆ABP is 38cm.
Find:-
Sides of the ∆ABP.
Assume:-
That perpendicular P lies on the segment AB at point C.
Solution:-
[ Refer the attachment for the figure. ]
AP exceeds the length AB by 7 cm.
According to question and figure -
- AP = BP = y
- AB = x
- AC = CB = x/2
According to the question,
→ y = x + 7 ....(1)
Also, given that the perimeter of the triangle is 38 cm.
The perimeter of the triangle is the sum of it's all sides.
i.e.
→ y + y + x = 38
→ 2y + x = 38
→ 2(x + 7) + x = 38 [From (1)]
→ 2x + 14 + x = 38
→ 3x = 24
→ x = 8
Substitute value of x = 8 in (eq 1)
→ y = 8 + 7
→ y = 15
As -
- AP = BP = y
- AB = x
So,
→ AP = BP = 15 cm
→ AB = 8 cm
•°• Sides of the triangle are 8cm, 15cm and 15cm
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